Two soccer players start from rest, 27 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.48 m/s2. The second player’s acceleration has a magnitude of 0.43 m/s2.

(a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?

To find the time it takes for the players to collide, we can use the equation of motion:

\[s = ut + \frac{1}{2}at^2\]

where:
s = distance traveled
u = initial velocity (which is 0 as both players start from rest)
a = acceleration
t = time

For the first player:
Given that the initial distance between the players is 27 m and the first player's acceleration is 0.48 m/s², we can use the equation:

\[27 = 0 + \frac{1}{2}(0.48)t^2\]

Simplifying the equation gives us:
\[27 = 0.24t^2\]

Dividing both sides of the equation by 0.24 gives us:
\[t^2 = \frac{27}{0.24}\]

\[t^2 = 112.5\]

Taking the square root of both sides gives us:
\[t = \sqrt{112.5} \approx 10.61\text{ s}\]

Therefore, it takes approximately 10.61 seconds for the players to collide.

To find how far the first player has run at the instant of the collision, we can use the equation of motion again. This time, we will use the time we calculated in the previous step (10.61 s) and the acceleration of the first player (0.48 m/s²).

\[s = ut + \frac{1}{2}at^2\]

Substituting the values, we get:
\[s = 0 \times 10.61 + \frac{1}{2}(0.48)(10.61)^2\]

Simplifying the equation gives us:
\[s = \frac{1}{2}(0.48)(112.5) \approx 27\text{ m}\]

Therefore, at the instant they collide, the first player has run approximately 27 meters.